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QUESTION: 1

is equal to

Solution:

We have:

limx→0 ln(1+ax)]/x=a

lim x→0 ln(1+ax)/ax=a

and then:

lim x→0 [ln(1+ax)-ln(1+bx)]/x

= a-b

QUESTION: 2

is equal to

Solution:

On rationalizing,

Lt x-->0 (1-Cosx)/[√(1+x) +1]

= lt x-->0 2sin^{2}(x/2)[√(1+x)+1] /x

= lt x-->0 [2sin(x/2)/x].sinx/2 [√(1+x) +1 ]

= 2.1/2 .0[1+1]=0

QUESTION: 3

Let f (x) = x sin 1/x, x ≠ 0, then the value of the function at x = 0, so that f is continuous at x = 0, is

Solution:

f(0) = lim(x→0) x sin (1/x)

We know ∀ x ∈ R,sin (1/x) ∈ [−1,1]

Hence, f(0) = lim(x→0) xsin(1/x)

= 0

QUESTION: 4

is equal to

Solution:

lim(x→0) (tanx−sinx)/x^{3}

=lim(x→0) [sinx(x−cosx)]/x^{3}cosx

=lim(x→0) (sinx)/x × (1−cosx)/x^{2} × (1/cosx)

= 1 × 1/2 × 1/1

= 1/2

QUESTION: 5

is equal to

Solution:

lim x→0 (1−cosx)/(xlog(1+x))

= lim x→0 {2sin^{2} x/2}/[xlog(1+x)]

= lim x→0 {{2sin^{2}x/2/(x/2)^{2}}*/1/4}/[1/xlog(1+x)]

= lim x→0 [log(1+x)/x]1/2

= 2

QUESTION: 6

is equal to

Solution:

QUESTION: 7

Solution:

Consider the given function.

lim x→0 ((1-x)^{n} −1)/x)

This is 0/0 form.

So, apply L-Hospital rule,

lim x→0 [n(1-x)^{n}−1 −0)]/1

lim x→ 0(n(1-x)^{n}−1)

=n(1-0)^{n}−1

= n - 1

QUESTION: 8

is equal to

Solution:

lim(x → 0) (tanx-x)/x^{2} tanx

As we know that tan x = sinx/cosx

lim(x → 0) (sinx/cosx - x)/x^{2}(sinx/cosx)

lim(x → 0) (sinx - xcosx)/(x^{2} sinx)

lim(x → 0) cosx - (-xsinx + cosx)/(x^{2}cosx + sinx (2x))

lim(x → 0) (cosx + xsinx - cosx)/x^{2}cosx + 2xsinx)

lim(x → 0) sinx/(xcosx + 2sinx)

Hence it is 0/0 form, apply L hospital rule

lim(x → 0) cosx/(-xsinx + cosx + 2cosx)

⇒ 1/(0+1+2)

= 1/3

QUESTION: 9

Solution:

lim x-->0 ((x^{2cosx-})/1-cosx)

Differentiating numerator and denominator

lim x-->0 (2xcosx - x^{2sinx})/sinx

Differentiating it for a second time

lim x-->0 (2cos - 2xsinx - 2xsinx - x^{2cosx})/cosx

Substituting x = 0

= lim x-->0 2/1

= 2

QUESTION: 10

is equal to

Solution:

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x^{3}.

Substiture x = 3y , so that, as x→0 , y→0.

∴ L = lim y→0 sin3y − 3y/(3y)^{3},

= lim y→0 (3siny − 4sin3y)− 3y)/27y^{3},

= lim y→0 {3(siny − y)/27y^{3}) − (4sin^{3}y/27y^{3})},

⇒ L = lim y→0 1/9 * (siny − y)/y^{3}) − 4/27* (siny/y)^{3}...(∗)

.Note that, here,

= lim y→0(siny − y)/y^{3})

= lim x→0 (sinx−x)/x^{3}) = L.

Therefore, (∗) ⇒ L = 1/9*L − 4/27

or, 8/9L = −4/27

Hence, L = −4/27*9/8

=−1/6

QUESTION: 11

is equal to

Solution:

QUESTION: 12

then for f to be continuous at x = 0, f (0) must be equal to

Solution:

QUESTION: 13

Solution:

QUESTION: 14

where a > 0, is equal to

Solution:

QUESTION: 15

is equal to

Solution:

QUESTION: 16

is equal to

Solution:

QUESTION: 17

is equal to

Solution:

lim(h → 0) [sin^{2}(x+h) - sin^{2}x]/h

l = lim(h→0) = [2sin(x+h)cos(x+h)]/1

{As we know that sin2x = 2sinxcosx

= sin2x

QUESTION: 18

If a is a real number, then

Solution:

QUESTION: 19

If k be a integer, then is equal to

Solution:

- lim x-->k+ = x-{x}

Lim x→k+ k-{k}

= 0

QUESTION: 20

is equal to

Solution:

**Correct Answer :- C**

**Explanation :** lt(x->a-) -|x-a|/x-a

= - |-a-a|/(-a-a)

= -|-2a|/(-2a)

= -2a/(-2a)

= 1

QUESTION: 21

is equal to

Solution:

QUESTION: 22

Maximum value of x^{3}−3x+2in [0,2] is

Solution:

QUESTION: 23

Solution:

√x / √(16+√x) - 4

= (√x * √(16+√x) + 4) / ( √(16+√x) - 4) * (√(16+√x) + 4)

= (√x * √(16+√x) + 4) / 16+√x - 16

= √(16+√x) + 4

now at lim x->0

= √16 + 4 = 8

QUESTION: 24

is equal to

Solution:

QUESTION: 25

The value of the limit Lim_{x→0} (cos x)cot^{2 x }is

Solution:

Given, Lim_{x→0} (cos x)^{cot² x}

= Lim_{x→0} (1 + cos x – 1)^{cot² x}

= eLim_{x→0} (cos x – 1) × cot² x

= eLim_{x→0} (cos x – 1) / tan² x

= e-1/2

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